A study considered risk factors for HIV infection among IV drug users. It found

Question

A study considered risk factors for HIV infection among IV drug users. It found that 40% of users who had <= 100 injections (light users) per month and 55% of users who had >100 injections (heavy users) per month were HIV positive.

1. Suppose we have a group of 10 light users and 10 heavy users. What is the probabiity that exactly 3 of the 20 users are HIV positive?

2. What is the probability that at least 4 of the 20 users are HIV positive?

3. Is the distribution of the number of HIV positive among the 20 users binomial? Why or why not?

Explanation / Answer

P(a light user is HIV positive ) = 0.4

P(a heavy user is HIV positive) = 0.55

P(3 out of 20 users are positive) = P(3 light users are HIV positive) +P(2light users and 1 heavy user is HIV positive)+ P( 1 light user and two heavy user are HIV positive) +P(3 heavy users are HIV positive)

P(3 out of 20 users are positive) = C(10,3)(0.4)3*(0.6)7(0.45)10 +C(10,2)(0.4)2*(0.6)8C(10,1)(0.45)9(0.55)+ C(10,2)(0.4)1(0.6)9C(10,1)(0.45)8(0.55)2+C(10,3)(0.6)10(0.55)3(0.45)7

P(3 out of 20 users are positive) = 0.002

2.

P(4 out of 20 users are HIV positive) = P(4 light users are HIV positive)+P(3light users and 1 heavy user are HIV positive)+P(2 light user and 2 heavy user are HIV positive)+P(1 light user and 3 heavy user are HIV positive)+P(4 heavy users are HIV positive)

P(4 out of 20 users are HIV positive) = Bin(10,4,0.4)Bin(10,0,0.55) + Bin(10,3,0.4)Bin(10,1,0.55) + Bin(10,2,0.4)Bin(10,2,0.55) + Bin(10,1,0.4)Bin(10,3,0.55)+Bin(10,0,0.4)Bin(10,4,0.55)

P(4 out of 20 users are HIV positive) = 0.008

3. The distribution is not binomial , but it is a combination of binomial.

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